Torque and Angular Momentum

Key idea: Understand angular momentum and torque, use τ = dL/dt for rotation, and apply conservation of angular momentum when external torque is negligible.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • show an understanding of and use the terms angular displacement, angular velocity, and angular acceleration of a rigid body with respect to a fixed axis
  • solve problems using the equations of motion for uniform angular acceleration that are analogous to the equations of motion for uniform linear acceleration
  • show an understanding of and use the terms angular momentum and moment of inertia of a rotating rigid body
  • calculate the moment of inertia about an axis for simple bodies by using calculus, the parallel-axis theorem or otherwise (knowledge of the perpendicular-axis theorem and mathematical derivation of the moment of inertia for spheres are not required)
  • show an understanding of torque produced by a force relative to a reference point, and apply the principle that torque is related to the rate of change of angular momentum to solve problems, such as those involving point masses, rigid bodies, or bodies with a variable moment of inertia e.g. an ice-skater
  • Apply Eₖ,rot = ½Iω² to the rotational kinetic energy of a rigid body.
  • Derive Eₖ,rot = ½Iω² for a rigid body from the equations of motion.
  • recall and apply the result that the motion of a rigid body can be regarded as translational motion of its centre of mass with rotational motion about an axis through the centre of mass to solve problems, including the use of F ⩽ µN for solid surfaces in no-slip contact (no distinction is made between the coefficient of static and kinetic friction)

Angular momentum is the rotational analogue of linear momentum. The central H3 idea is: ∑τₑₓₜ = dL/dt for components about a fixed axis. For a rigid body with constant I, this reduces to ∑τₑₓₜ = Iα. Here, “net torque” always means the resultant external torque about the stated axis.

1. Definitions (Must Know)

  • Angular momentum, L:
    • particle about an origin: vector L = vector r × vector p
    • fixed-axis rigid body (magnitude): L = Iω
  • Torque, τ about an axis/pivot:
    • magnitude: τ = rF sin φ = Fd_⊥
  • Torque–angular momentum relation: ∑τₑₓₜ = dL/dt
  • Symbols used in this lesson: vector r position vector (m), vector p momentum (kg m s⁻¹), vector L angular momentum (kg m² s⁻¹), τ torque (N m), I moment of inertia (kg m²), ω angular speed (rad s⁻¹), α angular acceleration (rad s⁻²).

Quick comparison:

SystemAngular momentumWhen to use
Particle about an originvector L = vector r × vector pPoint-mass / orbit-type setups
Fixed-axis rigid bodyL = IωMost H3 rigid-body rotation questions

2. Key Ideas (What Earns Marks)

  • Always specify the reference point/axis when talking about torque or angular momentum.
  • Torque changes angular momentum: no net external torque ⇒ angular momentum is conserved (about that axis).
  • For fixed-axis rigid bodies: L = Iω is the high-speed exam tool.
  • Variable I problems (ice skater): if τₑₓₜ ≈ 0, then Iω stays constant even though ω changes.

3. Detailed Explanations

A. Why the axis matters

Both τ and L are defined about a point/axis. Changing the axis changes:

  • the moment arm d_⊥ (hence the torque), and
  • the distance vector vector r (hence the angular momentum).

B. Linking torque and angular momentum (fixed-axis intuition)

For a point mass moving in a circle of radius r about an axis, the tangential force Fₜ changes the tangential speed, hence changes the “rotational motion”. The torque about the axis is: τ = rFₜ

For a rigid body about a fixed axis, all the parts rotate together with the same ω. The result you use is: ∑τₑₓₜ = dL/dt

If L = Iω and I is constant:

∑τₑₓₜ = d(Iω)/dt = Idω/dt = Iα

This connects directly to the previous two lessons:

C. When I changes (ice skater idea)

If there is negligible external torque about the spin axis, then dL/dt ≈ 0, so L is approximately constant.

For a spinning skater, you can model L ≈ Iω. Pulling arms in reduces I, so ω increases to keep Iω constant.

4. Common Mistakes

  • Writing “angular momentum is conserved” without stating “when external torque is negligible (about the chosen axis)”.
  • Using L = Iω for a particle not in rigid rotation about the axis (use vector L = vector r × vector p if needed).
  • Forgetting that τ = Fd_⊥ depends on the perpendicular distance to the line of action (not just “the radius”).
  • Mixing up L (angular momentum) with I (moment of inertia) because both appear in rotational formulas.

5. Exam Tips

  • If the question screams “conservation”, check the external torque about the relevant axis:
    • Is there a pivot reaction through the axis? (often gives zero torque)
    • Is friction/air resistance negligible? (often assumed)
  • Use the reasoning chain: negligible external torque ⇒ constant L ⇒ constant Iω for fixed-axis rotation.
  • For fixed-axis questions, start from L = Iω and ∑τₑₓₜ = dL/dt; it’s faster than re-deriving.

6. Worked Examples

Modelled example 1

Net torque gives rate of change of angular momentum

Core

Problem

A flywheel experiences constant net torque 1.5 N m for 4.0 s. Find |Δ L|.
Study the worked solution
  1. Use angular impulse

    Method

    Δ L = τₙₑₜΔ t.

    Reason

    Integrating dL/dt = τₙₑₜ over constant torque gives the product.

    Working

    Δ L = ∫τ dt
  2. Evaluate

    Method

    |Δ L| = 6.0 kg m² s⁻¹.

    Reason

    Multiply torque by duration.

    Working

    (1.5)(4.0) = 6.0 kg m² s⁻¹

Guided practice 2

Ice skater (variable moment of inertia)

About 5 min

Problem

A skater changes from I₁ = 3.0 kg m², ω₁ = 2.0 rad s⁻¹ to I₂ = 1.2 kg m² with negligible external torque. Find ω₂.

Try this before viewing the solution

Hints

Hint 1: check the external torque
Negligible external torque means angular momentum about the axis is conserved.
View solution step by step
  1. Conserve angular momentum

    Method

    I₁ω₁ = I₂ω₂.

    Reason

    τₑₓₜ ≈ 0.

    Working

    L₁ = L₂
  2. Solve

    Method

    ω₂ = 5.0 rad s⁻¹.

    Reason

    Reducing inertia requires increased angular speed for fixed L.

    Working

    (3.0)(2.0)/1.2 = 5.0

Common misconception 3

Fixed-axis rigid body: torque and angular acceleration

Find and correct the mistake

Learner claim

For I = 0.60 kg m² and net torque 2.4 N m, a learner says both α and dL/dt equal 4.0. Explain what has been confused and calculate both quantities.

Try this before viewing the solution

Rate of angular-momentum change

View solution step by step
  1. Find angular acceleration

    Method

    α = 4.0 rad s⁻².

    Reason

    For constant I, τ = Iα.

    Working

    α = 2.4/0.60 = 4.0
  2. Find momentum rate

    Method

    dL/dt = 2.4 kg m² s⁻².

    Reason

    The general law is τₙₑₜ = dL/dt.

    Working

    dL/dt = τₙₑₜ = 2.4

Examiner practice 4

Finding L from I and ω

3 marks

Examination question

A turntable has I = 0.40 kg m² and ω = 12 rad s⁻¹. Find L. [3 marks]

Try this before viewing the solution

View solution step by step
  1. State relation

    1 mark

    Method

    L = Iω.

    Reason

    The body rotates rigidly about the stated fixed axis.

    Working

    L = Iω
  2. Substitute

    1 mark

    Method

    L = (0.40)(12).

    Reason

    Use SI quantities.

    Working

    L = 4.8
  3. Report

    1 mark

    Method

    L = 4.8 kg m² s⁻¹.

    Reason

    These are angular-momentum units.

    Working

    4.8 kg m² s⁻¹

Challenge 5

Angular impulse from a changing torque (area-under-graph idea)

Minimal support

Independent transfer

Net torque varies as τ(t) = 3.0t in SI units from 0 to 2.0 s. Find |Δ L|.

Try this before viewing the solution

Hints

Hint 1: use area under torque-time graph
Angular impulse is ∫τ dt.
View solution step by step
  1. Set up the integral

    Method

    Δ L = ∫₀² 3.0t dt.

    Reason

    Torque is the time derivative of angular momentum.

    Working

    Δ L = ∫τ dt
  2. Evaluate

    Method

    Δ L = 6.0 kg m² s⁻¹.

    Reason

    The integral equals the triangular area under the torque-time line.

    Working

    3.0[t²/2]₀² = 6.0

7. Mind Stretchers

Mind stretcher 1: Why a figure skater speeds up (concept + energy)Extension

If I decreases and L is conserved, ω increases. Where does the extra rotational kinetic energy come from?

Answer

From work done by the skater’s muscles while pulling the arms inward (internal forces can do work on parts of the body, increasing rotational kinetic energy even though external torque is negligible).

Mind stretcher 2: “Conserved about which axis?”Extension

An object is rotating while a force acts on it. How do you decide whether angular momentum is conserved in an exam question?

Answer

Pick the axis or reference point, then check the external torque about it. If τₑₓₜ ≈ 0, angular momentum is conserved about that axis. If there is a significant external torque about your chosen axis, angular momentum about that axis is not conserved.

8. Optional/Enrichment: Vector Form and 3D Rotation

The full relation is vector-based: ∑ vector τₑₓₜ = (d vector L)/dt with vector L = vector r × vector p. For 3D rigid body rotation (changing axis direction), you need more advanced dynamics; H3 fixed-axis questions typically avoid this unless explicitly guided.

Next step

Return to the Rotational Motion hub, or continue in sequence to Rotational Kinetic Energy.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027